Astronomy Assignment #9 Solutions
Text Problems:
Unit Number
56
57
58
Questions for Review
1, 4, 6
5, 6, 7
1, 2, 3, 4, 5
Problems
1, 2, 6
1, 2, 3, 4, 7,
8
1, 2, 4, 6
Test Yourself
Problems in Unit 56
1.
Two stars are in a binary system. Find there combined mass if they have
a. an orbital period, P, of 5 years and an orbital separation of, a, of 10 AU.
a 3AU (10 AU) 3

 40 Solar masses
Pyr2
(5 yr) 2
The combined mass of the binary star system is 40 solar masses.
Use the combined mass relation, m A  mB 
b. an orbital period, P, of 2 years and an orbital separation of, a, of 4 AU.
a 3AU (4AU) 3

 16 Solar masses
Pyr2
(2 yr) 2
The combined mass of the binary star system is 16 solar masses.
Use the combined mass relation, m A  mB 
2. Mizar A and B are about 400 AU apart currently. Their orbital period is estimated to be
between 2,000 and 4,000 years. What range of masses does this suggest for the pair?
Use the combined mass relation twice to bracket the possible values,
a3
(400AU) 3
m A  mB  AU2 
 16 Solar masses
Pyr
(2,000 yr) 2
a 3AU
(400AU) 3
m A  mB  2 
 4 Solar masses
Pyr
(4,000 yr) 2
The combined mass of the binary star system is between 4 and 16 solar masses
6. Two stars are orbiting each other, both 4.2 arcsec from their center of mass. Their orbital
period is 420.3 years and their distance from the Earth is 104 ly. Find their masses.
This problem requires you to first use the angular size formula to calculate the true length from
each star to their center of mass from angular separation (size) of 4.2”. After these lengths are
known, then the semi-major axis, a, is given by the sum of those two lengths (see page 462).
Once the semi-major axis, a, is known and the orbital period is known it is simple to calculate the
combined mass of the star system (see pg 461). Since both stars are the same distance from the
center of mass, then the stars have equal masses. Thus, the mass of each star will be half the
mass of the combined mass of the star system.
First calculate the true length from each star to their center of mass from angular separation
knowing the angular separation of 4.2” = 1.167x10-3 degrees. Since we need the true length, L,
in units of AU, we will convert 104 ly into 6.577x106 AU.
A
L


2D
360
A
L  2D 
360 

 1.16736010 
3 
L  2 6.577 10 6 AU 

 134 AU
The semi-major axis, a, of the stars is 268 AU.
Second apply Kepler’s Third Law
268AU
a 3AU
MA  MB  2 
 109 M Sun
Pyr
420.32yr
3
Thus, the combined mass of the star system is 109 solar masses. Since the stars have equal mass,
as indicated by their equal angular distances from the center of mass, the mass of each star is
54.5 solar masses. If these stars were main sequence stars they would have to be O stars with
such a high mass.
Problems in Unit 57
1. Alpha Centauri is actually two stars that orbit each other. Recent measurements show that
Alpha Centauri A has an angular diameter of 0.0085 arcsec, while Alpha Centauri B has an
angular diameter of 0.0060 arcsec. The pairs are at a distance of 4.4 lyr. What are their
diameters compared to the Sun.
This is a fairly messy problem. There is
one short cut we can use. After we calculate the radius of Alpha Centauri A, we can simply calculate
the radius of Alpha Centauri B using the ratio of their angular sizes, Alpha Centauri B will be a factor of
(60/85) smaller than Alpha Centauri A.
To determine the angular size of Alpha Centauri A we need to use the angular size formula
A
L

, where A is the angular diameter of the star, L is its true diameter, and D is its distance.

2D
360
We need to solve for L. We will have to change 4.4 light years into meters and 0.0085 arcsec into
degrees to get a meaningful answer.
A
L


2D
360

 1  
 .0085"

 3,600"  





A
m


 L  2D 
 2   4.4 ly   9.46  1015    


ly  
360
360



L  2  4.16  10
16
2.36  10 6
m
 1.71  10 9 m

360
The diameter of Alpha Centauri A is 1.71 x 109 meters. The Sun’s diameter is 1.39 x 109 meters
as determined from the table in the text’s appendix. Thus, Alpha Centauri A is slightly larger
than the Sun with a diameter of 1.23 solar diameters.
Alpha Centauri B is (60/85) = 0.706 times smaller than Alpha Centauri A. based on the ratio of
their angular sizes (and the fact that they are at the same distance). So Alpha Centauri B is
slightly smaller than the Sun with a diameter of 0.867 solar diameters.
2. A star with the same color as the Sun is found to produce a luminosity 81 times larger. What is
its radius, compared to the Sun’s?
The following three problems all use the Stefan-Boltzmann Law L  4R 2  T 4 . We can avoid
painful and error prone calculations, by using a ratio technique as illustrated below.
2
4
2
4
LStar  4RStar
 TStar
and LSun  4RSun
 TSun
Take the ratio of the two equations above…
2
4
R
LStar 4RStar
 TStar

  Star
2
4
LSun 4RSun  TSun  RSun
L
  Star
 LSun
  RStar
  
  RSun



2
T
  Star
 TSun






2
T
  Star
 TSun



4
4
Now we have an expression where only ratios are used…much simpler that working with the full
equation. This problem gives us the ratio of temperatures (1, since the stars are the same color
they are also the same temperature) and the ratio of the luminosities (81). We need to solve for
the ratio of radii.
L
  Star
 LSun
  RStar
  
  RSun
2
  TStar
  
  TSun



4
 LStar 


LSun  81
 RStar 


 
 4  81
4
1
 TStar 
 RSun 


 TSun 
The ratio of the radii squared is 81, so the ratio of the radii must be 81  9 . The radius of the
star in question is 9 times the radius of the Sun. A note is worthy here: We expected this star to
be larger because it was the same temperature as the Sun, by quite a bit more luminous. What
this problem is trying to illustrate, is the luminosity depends on the square of the stellar radius,
not just on the radius alone.
2
3. A star with the same radius as the Sun is found to produce a luminosity 81 times larger. What
is its surface temperature compared to the Sun?
As in the preceding question
L
  Star
 LSun
 TStar

 TSun
  RStar 
  

  RSun 
 LStar 


4
LSun 


 
2
 RStar 



 RSun 
2
T
  Star
 TSun




4
81  81
12
TStar 4
 81  3
TSun
This star is only three times hotter than the Sun. Notice how important temperature is to
luminosity. A factor of three increase in temperature increases the luminosity by a factor or 81!

4. The surface temperature of Arcturus is about half as hot as the Sun’s, but Arcturus is about
100 times more luminous that the Sun. What is its radius, compared to the Sun’s?
As before
L
  Star
 LSun
  RStar
  
  RSun



2
T
  Star
 TSun



4

100  100  1,600
 
4
0.0625
1

 
2
The ratio of the radii squared is 1600, so the ratio of the radii must be 1,600  40 . The radius
of Arcturus is 40 times the radius of the Sun. A note is worthy here.
 RStar

 RSun
2
7. A star is five times as luminous as the Sun and has a surface temperature of 9,000 K. What
is its radius compared to that of the Sun?
This is a classic Stefan-Boltzmann Problem.
2
4
2
4
LStar  4RStar
 TStar
and LSun  4RSun
 TSun
Take the ratio of the two equations above…
2
4
R
LStar 4RStar
 TStar

  Star
2
4
LSun 4RSun  TSun  RSun
L
  Star
 LSun
  RStar
  
  RSun
2
  TStar
  
  TSun
2
R
5   Star
 RSun
  9,000 K 
  

  5,800 K 
R
5   Star
 RSun

  1.554

R
5   Star
 RSun

  5.80


2
  TStar
  
  TSun






4
4
4
2
2
RStar
5

 .928
RSun
5.80
The star has a radius just slightly smaller than the Sun at 0.928 solar radii.
8. Using the information that Betelgeuse has an angular size of 0.05 arcsec and is 130 pc from
Earth, show how you can calculate its diameter in km.
A
L
, where A = 0.05” = 1.39x10-5, and


2D
360
D = 130 pc = 130 x (3.08x1016m) = 4.00x1018m.
This is an angular size problem. So we use
A
L


2D
360



A
1.38  10 5
L

2

D

 2  (4.00  1018 m)  9.63  1011 m


360
360
The diameter of Betelgeuse is 9.63 x 1011 m or 692 solar diameters.
Problems in Unit 58
1. Estimate the luminosity and temperature of the star Acrux by examining the HR diagram
in Figure 58.3. Using the values you estimated, determine the size of Acrux as compared to
the Sun, using the Stefan-Boltzmann Law.
According to Figure 58.3, Has a luminosity of about 3x104 LSun and a temperature of about
30,000K. It’s radius compared to the Sun is most easily found the technique in problem &
above.
2
4
R
LStar 4RStar
 TStar

  Star
2
4
LSun 4RSun  TSun  RSun
L
  Star
 LSun
  RStar
  
  RSun
2
  TStar
  
  TSun
2
R
3  10   Star
 RSun
  30,000 K 
  

  5,800 K 
R
3  10   Star
 RSun

  5.17 4

R
3  10   Star
 RSun

  716

4
4
4
2
  TStar
  
  TSun






4
4
4
2
2
RStar
3  10 4


 6.47
RSun
716
Acrux has a radius of about 6.47 times the radius of the Sun. This is consistent with the radius
read from the HR diagram in Figure 58.4
2. Alnitak is the brightest O-class star in the night sky, located at the eastern end of Orion’s
belt. Proxima Centauri is the nearest M-class main sequence star. Estimate the
luminosities and temperature of these two stars from their positions on the HR diagram
(Fig 58.3). Calculate the ratio of their sizes using the Stefan-Boltzmann law.
From examining the H-R diagram in Figure 58.4 I estimate the following parameters for the two
stars
Star
Alnitak
Proxima Centauri
Luminosity, LSun
2 x 105
2 x 10-4
Temperature,
K
30,000
3,000
The Stefan-Boltzmann Law will yield a ratio of the radii of the stars as appears below
L A  TA4  4R A2
4
2
LPC  TPC
 4RPC
LA
TA4  4R A2

4
2
LPC TPC
 4RPC
LA
T 4  R2
 4A 2A
LPC TPC  RPC
L A  TA

LPC  TPC
 2  10 5 L Sun

4
 2  10 L Sun



4
 R
  A
 RPC



  30,000 K 
  

  3,000 K 
 R
1 10  1 10   A
 RPC
9
 RA

 RPC
R
 A 
RPC
4
2
4
 R
  A
 RPC



2



2
2

  110 5

110 5  316
Thus, Alnitak is approximately 316 times larger in radius compare to Proxima Centauri.
4. If a main sequence star has a luminosity of 3,000 LSun, what is its mass in relation to the
Sun’s?
Use the mass-luminosity relation LM3.5. The mass of this main sequence star is given
1
1
approximately as M  L3.5  (3,000) 3.5  9.85 solar masses.
6. How massive would a main sequence star need to be, to be 100 times brighter (more
luminous) than the Sun?
Use the mass-luminosity relation LM3.5. The mass of this main sequence star is given
1
3.5
approximately as M  L
 (100)
1
3.5
 3.73 solar masses.
Instructor Assigned Topic: Write a quantitative critical comparison between the twenty nearest star
systems in the sky and the Sun. A list of properties of the twenty nearest star systems is attached.
Answer the question “How are the nearest stars in the sky like or unlike the Sun?” Does the Sun appear
to be a common type of star in the set nearest stars? Be quantitative in your response by averaging
appropriate properties of the twenty nearest star systems and discuss the meanings of the average
properties.
Extra credit awarded for including HR diagrams of the stars in the twenty nearest star systems and in the
twenty brightest stars. A short paragraph interpreting the HR diagram must accompany the diagram for
credit.
The Twenty Nearest Star Systems
Common Name
1
2
3
4
5
Alpha Centauri A
Alpha Centauri B
Proxima Centauri
Barnard's Star
Wolf 359
Lalande 21185
Sirius
Sirius B
6
7
8
9
10
11
UV Ceti
Ross 154
Ross 248
epsilon Eridani
Lacaille 9352
Ross 128
EZ Aquarii
12
13
Procyon
14
61 Cygni
15
16
17
18
19
20
epsilon Indi
DX Cancri
tau Ceti
RECONS 1
Average Values
Catalog
Name
# of
Stars
Spectral
Type
Luminosity
Class
Gl 559 A
Gl 559 B
Gl 551
Gl 699
Gl 406
Gl 411
Gl 244 A
Gl 244 B
Gl 65 A
Gl 65 B
Gl 729
Gl 905
Gl 144
Gl 887
Gl 447
Gl 866 A
Gl 866 B
Gl 866 C
Gl 280 A
Gl 280 B
Gl 820 A
Gl 820 B
Gl 725 A
Gl 725 B
Gl 15 A
Gl 15 B
Gl 845
GJ 1111
Gl 71
GJ 1061
30 Stars
3
G2
K0
M5.5
M4.0
M6.0
M2.0
A1
DA2
M5.5
M6.0
M3.5
M5.5
K2
M1.5
M4.0
M5.0
V
V
V
V
V
V
V
V
V
V
V
V
V
V
V
V
1
1
1
2
2
1
1
1
1
1
3
2
2
2
2
1
1
1
1
1 - 12
2-6
3-2
---F5
DA
K5.0
K7.0
M3.0
M3.5
M1.5
M3.5
K5
M6.5
G8
M5.5
O-0
B-0
A-1
F-1
G-2
K-5
IV-V
V
V
V
V
V
V
Ve
V
Vp
V
V - 26
IV - 1
Apparent
Magnitude,m
Absolute
Magnitude,
M
Distance,
lyrs
0.01
1.34
11.09
9.53
13.44
7.47
-1.43
8.44
12.54
12.99
10.43
12.29
3.73
7.34
11.13
13.33
13.27
14.03
0.38
10.7
5.21
6.03
8.9
9.69
8.08
11.06
4.69
14.78
3.49
13.03
8.57
4.38
5.71
15.53
13.22
16.55
10.44
1.47
11.34
15.4
15.85
13.07
14.79
6.19
9.75
13.51
15.64
15.58
16.34
2.66
12.98
7.49
8.31
11.16
11.95
10.32
13.3
6.89
16.98
5.68
15.21
11.26
4.36
4.40
4.22
5.96
7.78
8.29
8.58
8.58
8.72
8.72
9.68
10.32
10.52
10.74
10.91
11.26
11.26
11.26
11.40
11.40
11.40
11.40
11.52
11.52
11.62
11.62
11.82
11.82
11.88
11.92
9.83
M - 17
A summary of the nearest stars as indicated by the twenty nearest star systems should contain the
following points
 The 20 nearest star systems contain 30 stars, only 40% (12/30) of the stars are solitary stars like
the Sun,
 85% (22/26) of the stars appear to be a cooler spectral class than the Sun,
 They are almost all main sequence dwarf stars like the Sun,
 On average they are invisible with average m = 8.57, and
 On average they are about 6 magnitudes less luminous than the Sun, or about 1/300 the Sun’s
luminosity.
 There is one star, Alpha Centauri, that is similar to the Sun.

HR Diagram
Main Sequence Stars
-10
Absolute Magnitude
-5
0
5
10
15
20
O0 O5 B0 B5 A0 A5 F0 F5 G0 G5 K0 K5 M0 M5
Spectral Type